Explore topic-wise InterviewSolutions in .

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

1.

`tan^(-1)[(cosx)/(1+sinx)]`is equal to`pi/4-x/2,forx in (-pi/2,(3pi)/2)``pi/4-x/2,forx in (-pi/2,pi/2)``pi/4-x/2,forx in (-pi/2,pi/2)``pi/4-x/2,forx in (-(3pi)/2,pi/2)`

Answer» `tan^-1((cosx)/(1+sinx))`
`=tan^-1((sin(pi/2-x))/(1+cos(pi/2-x)))`
`=tan^-1((2sin(pi/4-x/2)cos(pi/4-x/2))/(2cos^2(pi/4-x/2)))`
`=tan^-1(tan(pi/4-x/2)`
`=pi/4-x/2`
Now, we know, range of `tan^-1x` is from `-pi/2` to `pi/2`.
`:. -pi/2 lt pi/4-x/2 lt pi/2`
`=> -(3pi)/4 lt -x/2 lt pi/4`
`=> -pi/2 lt x lt (3pi)/2`
So, option `a` is the correct option.
Discussion
2.

The value of `sin^(-1)(sin12)+sin^(-1)(cos12)=`

Answer» `sin^-1(sin12)+cos^-1(cos12)`
Here, we have to convert `sin12` into a value from `-pi/2` to ` pi/2` and `cos12` into a value from `0` to `pi`.
We can write,
`sin^-1(sin12)+cos^-1(cos12) = sin^-1(sin(12-4pi))+cos^-1(cos(4pi-12))`
`=12-4pi+4pi-12 = 0`
`:. sin^-1(sin12)+cos^-1(cos12) =0`
Discussion
3.

Find the principal value of `sin^(-1)(-1/sqrt2)-2tan^(-1)(-sqrt3)+cos^(-1)(-1/2)`.

Answer» Correct Answer - `(13pi)/12`
Discussion
4.

Find the principal value of the following: (i) `cosec^(-1)(2)` (ii) `tan^(-1) (-sqrt3)` (iii) `cos^(-1) (-(1)/(sqrt2))`

Answer» (i) Let `cosec^(-1) (2) = y`. Then,
`cosec y = 2 = cosec ((pi)/(6))`
We know that the range of the principal values branch of the function `cosec^(-1) " is " [-(pi)/(2), (pi)/(2)] - {0} and cosec (pi)/(6) = 2`
Therefore, the principal value of `cosec^(-1) (2) " is " (pi)/(6)`
(ii) Let `tan^(-1) (-sqrt3) = y`. Then,
`tan y = -sqrt3 = - tan. (pi)/(3) = tan.(-(pi)/(3))`
We know that the range of the principal values branch of the function `tan^(-1) " is " (-(pi)/(2), (pi)/(2)) and tan (-(pi)/(3)) = -sqrt3` ltrbgt Therefore, the principal value of `tan^(-1)(-sqrt3) " is " -(pi)/(3)`
(iii) Let `cos^(-1) (-(1)/(sqrt2)) = y`. Then, `cos y = -(1)/(sqrt2) = - cos. ((pi)/(4)) = cos (pi - (pi)/(4)) = cos. ((3pi)/(4))`
We know that the range of the principal values branch of the function `cos^(-1) " is " [0, pi] and cos. ((3pi)/(4)) = -(1)/(sqrt2)`
Therefore, the principal value of `cos^(-1) (-(1)/(sqrt2)) " is " (3pi)/(4)`
Discussion
5.

The value of the expression `sin^(-1)(sin(22pi)/7)+cos^(-1)(cos(5pi)/3)+tan^(-1)(tan(5pi)/7)+sin^(-1)(cos2)`is(a) `(17pi)/(42)-2`(b) `-2`(c)`(-pi)/(21)-2`(d) `non eoft h e s e`

Answer» `sin^-1(sin((22pi)/7)) + cos^-1(cos((5pi)/3))+tan^-1(tan((5pi)/7)) + sin^-1(cos2)`
`=sin^-1(sin(3pi+pi/7)) + cos^-1(cos(2pi-pi/3)+tan^-1(tan(pi-(2pi)/7)) + pi/2-cos^-1(cos2)`
`=-sin^-1(sin(pi/7)) + cos^-1(cos(pi/3))-tan^-1(tan((2pi)/7)) + pi/2-2`
`=-pi/7+pi/3-(2pi)/7+pi/2 - 2`
`=(17pi)/42 - 2`
So, option `a` is the correct option.
Discussion
6.

The domain of definition of f(X) = `sin^(-1)(-x^(2))` isA. `[-1,1]`B. `[0,1]`C. `[-1,0]`D. `[-2,23]`

Answer» The domain of `sin^(-1)x is [-1,1]` Therefore f(x)=`sin^(-1)(-x^(2))` is defined for all x satisfying
`-1le-x^(2)le1`
`rarr 1gex^(2)-1rarr 0lex^(2)le1rarrx^(2)le1rarr x^(2)-1le0`
`rarr (x-1)(x+1)le0rarr-1lexle1`
Hence the domain of f(X)=`sin^(-1)(-x^(2))` is [-1,1]
Discussion
7.

If `x , y , z in [-1,1]`such that `sin^(-1)x+sin^(-1)y+sin^(-1)z=-(3pi)/2,`find the value of `x^2+y^2+z^2dot`A. 1B. 3C. `(3pi^(23))/(4)`D. `3pi^(2)`

Answer» We know that the minimum value of `sin^(-1) x for x in[-1,1]` is `-(pi)/(2)`
`therefore sin^(-1)xge-(pi)/(2)sin^(-1)yge-(pi)/(2) and sin^(-1)z ge -(pi)/(2)` for all
`rarr sin^(-1)x+sin^(-1)y+sin^(-1)zge(pi)/(2)+(pi)/(2)+(pi)/(2)`
` therefore sin^(-1)x+sin^(-1)y+sin^(-1)z=(3pi)/(2)`
`rarr sin^(-1)x=(pi)/(2)sin^(-1)y -(pi)/(2),sin^(-1)z=(pi)/(2)`
`rarr x==z=-1`
Hence `x^(2)+y^(2)+z^(2)=(1)^(2)+(-1)^(2)+(-1)^(2)=3`
Discussion
8.

The domain of definition of `cos^(-1)(2x-1)` isA. `[-1,1]`B. `[0,1]`C. `[-1,0]`D. `[0,2]`

Answer» The domain of `cos^(-1)x` is[-1,1] so the domain of
`cos^(-1)(2x-1)` is the set of all values of x satisfying
`-1le2x-1le1rarr0le2xle2rarr0lexle1`
Hence the domain of `cos^(-1)(2x-1)is [0,1]`
Discussion
9.

The value of `cos^(-1)(cos(5pi)/3)+sin^(-1)(sin(5pi)/3)`is`pi/2`(b) `(5pi)/3`(c) `(10pi)/3`(d) 0

Answer» Correct Answer - A
Discussion
10.

Find the principal value of `sin^(-1)(-1/sqrt2)+tan^(-1)(-1/sqrt3)+sec^(-1)(2)`

Answer» Correct Answer - `-pi/12`
Discussion
11.

The number of ordered triplets `(x,y,z)` satisfy the equation `(sin^(- 1)x)^2=(pi^2)/4+(sec^(- 1)y)^2+(tan^(- 1)z)^2`A. 2B. 4C. 6D. 8

Answer» Correct Answer - A
`(sin^(-1)x)in[-(pi)/(2),(pi)/(2)]`
`therefore (sin^(-1)x)^(2)le (pi^(2))/(4)`
`(sec^(-1)y)^(2), (tan^(-1)z)^(2)ge 0`
`therefore R.H.S. ge (pi^(2))/(4)`
`therefore (sin^(-1)x)^(2)=(pi^(2))/(4)`
`therefore (sex^(-1)y)^(2)+(tan^(-1)z)^(2)=0`
or `sec^(-1)y=tan^(-1)z=0`
`therefore sin^(-1)x = pm(pi)/(2), y = 1, z = 0`
Discussion
12.

If `x , y , z`are natural numbers such that `cot^(-1)x+cot^(-1)y=cot^(-1)z`then the number of ordered triplets `(x , y , z)`that satisfy the equation is0 (b) 1(c) 2 (d) Infinite solutions

Answer» Correct Answer - D
`x = 1, y =1` is not a solution of the given equation.
Suppose `(x, y) != (1,1)`
Then `((1)/(x) + (1)/(y))/(1-(1)/(xy)) = (1)/(z)`
`rArr x(z -y) = -(1 + yz)`
`rArr x = -((1 + xyz))/(z -y)`
If `y = n + 1, z = n " then " x = n^(2) + n+1`
All such numbers are solutions of the given equation
Discussion
13.

Absolute value of sum of all integers in the domain of `f(x)=cot^(-1)sqrt((x+3)x)+cos^(-1)sqrt(x^2+3x+1)`is_______

Answer» Correct Answer - `-3`
We must have `x(x + 3) ge 0`
`rArr x ge 0 " or " x le -3`...(i)
Also, `-1 le x^(2) + 3x + 1 le 1`
`rArr x(x + 3) le 0 rArr -3 le x le 0`...(ii)
From Eqs. (i) and (ii), we get `x = {0, -3}`
Also, for these values `x^(2) + 3x + 1 ge -1`
Hence, the required sum is `-3`
Discussion
14.

The domain of definition of f(x)=`sin^(-1)sqrt(x-1)`A. `[-1,1]`B. `[0,1]`C. `[,2]`D. `[2,3]`

Answer» The domain of `sin^(-1)is [-1,1]` so the domain of
`f(*x)=sin^(-1)sqrt(x-1)` is the set of values of x satisfying
`-1lesqrt(x-1)le1`
`rarr0lesqrt(x-1)le1`
`rarr0lex-1le1rarrlexle2rarr x in [1,2]`
Hence the domain of f(x)=`sin^(-1)sqrt(x-1)` is [1,2]
Discussion
15.

The domain and range of `f(x) = sin^1x + cos^-1 x +tan^-1x + cot^-1x + sec^-1x + cosec^-1x` respectively areA. `{-1,1},(3pi)/2`B. `{-1,1},pi/2`C. `(-1,1),pi/2`D. `(-1,1),2pi`

Answer» Correct Answer - A
Discussion
16.

If M and m are the greatest and least value of the function `f(x)=(cos^(-1)x)^2+(sin^(-1)x)^2` then the value of `((M+9m)/m)^3` is …..

Answer» Correct Answer - 1331
Discussion
17.

The value of `sin^(-1){cot(sin^(-1)(sqrt((2-sqrt3)/4)+cos^(-1)(sqrt(12)/4)+sec^(-1)sqrt2)}` is

Answer» Correct Answer - 256
Discussion
18.

The total number of ordered pairs of (x,y) satisfying the equation `13+12[tan^-1x]=24[In x]+8[e^x]+6[cos^-1y]` is/are:

Answer» Correct Answer - A
Discussion
19.

Total number of ordered pairs (x, y) satisfying `Iyl = cos x and y = sin^(-1) (sin x)` where `|x| leq3pi` is equal to

Answer» Correct Answer - 5
Discussion
20.

The domain of the function given by `f(x)=sqrt(sin^(-1)(2x)+pi/6)` isA. `[-1/2,1/2]`B. `[-1/4,1/2]`C. `[-1,1]`D. `[-1,1/2]`

Answer» Correct Answer - B
Discussion
21.

If `ax + b sec(tan^-1 x) = c and ay + b sec(tan^-ly) = c,` then `(x+y)/(1-xy)` is equal to

Answer» Correct Answer - `(2ac)/(a^(2) - c^(2))`
Discussion
22.

If `sin^(-1)x+sin^(-1)(1-x)+cos^(-1)x=0`, then x is

Answer» Correct Answer - D
Discussion
23.

If `tan^(-1)x+tan^(-1)y+tan^(-1)z=pi`, then `1/(xy)+1/(yz)+1/(zx)=`

Answer» Correct Answer - B
Discussion
24.

If xy +yz+zx=1 then `tan^(-1)x+tan^(-1)y+tan^(-1)z`=A. `(pi)`B. `pi//2`C. 1D. none of these

Answer» Correct Answer - B
Discussion
25.

`tan^(-1)x-tan^(-1)y=tan^(-1)((x-y)/(1+xy))` holds good forA. All `x, y in R`B. `absx lt 1absy lt 1`C. `absx gt1,absygt1`D. `xy gt -1`

Answer» Correct Answer - D
Discussion
26.

The set of vlues of x for which `tan^(-1)(x)/sqrt(1-x^(2))=sin^(-1)` x holds isA. RB. `(-1,1)`C. `(0,1)`D. `[-1,0]`

Answer» We observe that RHS is defined forall `x in [-1,1]` whereas LHS is meaningfull for -`1ltxlt1`
Also for `x in (-1,1)` we have
`tan^(-1)(x)/sqrt(1-x^(2))`
`theta=sin^(-1)x`
Discussion
27.

If A=`tan^-1((xsqrt3)/(2k-x))` and B=`tan^-1((2x-k)/(ksqrt3))` then find the value of `A-B` (A) `0^@` (B) `30^@` (C) `60^@` (D) `45^@`A. `0^(@)`B. `45^(@)`C. `60^(@)`D. `30^(@)`

Answer» Correct Answer - D
Discussion
28.

`sin^(-1) (5/x) +sin^(-1) (12/x) =pi/2`

Answer» `sin^(-1)(5/x)=pi/2-sin^(-1)(12/x)`
`pi/2-cos^(-1)(5/x)=pi/2-sin^(-1)(12/x)`
`cos^(-1)(5/x)=sin^(-1)(12/x)=theta`
`costheta=5/x,sintheta=12/x`
`sin^2theta+cos^2theta=1`
`144/x^2+25/x^2=1`
`169/x^2=1`
`x=pm13`
possible value is +13.
Discussion
29.

In a triangle ABC, if `2b=a+c` and `A-C=90`, then `sin B` equals

Answer» As, 2b=a+c
So by sine rule,`2sinB=sinA + sinC`
`2sinB=2sin((A+C)/2)cos((A-C)/2)``sinB=sin((pi-B)/2)cos(pi/4)``2sin(B/2)cos(B/2)=cos(B/2)*(1/sqrt2)``sin(B/2)=1/(2sqrt2)`
Hence, `cos(B/2)=sqrt7/(2sqrt2)`
So, `sin(B)=2*sin(B/2)*cos(B/2)=2*sqrt7/(4*2)=sqrt7/4`
Discussion
30.

If `tan^(-1)(x+3/x)-tan^(-1)(x-3/x)=tan^(-1)6/x ,`then the value of `x^4`is_____.

Answer» Correct Answer - 9
`tan^(-1) (x + (3)/(x)) - tan^(-1) (x -(3)/(x)) = tan^(-1).(6)/(x)`
or `tan^(-1) (((x + (3)/(x)) - (x -(3)/(x)))/(1 + (x + (3)/(x)) (x -(3)/(x)))) = tan^(-1).(6)/(x)`
`rArr x^(2) -(9)/(x^(2)) = 0 " or " x^(4) = 9`
Discussion
31.

The graph of inverse trigonometric function can be obtained from the graph of their corresponding function by interchanging X and Y-axes.

Answer» Correct Answer - 1
We know that, the graph of an inverse function can be obtained from the corresoponding graph of original function as a mirror image (i.e., reflection) along the line `y = x`.
Discussion
32.

Solve :`cos^(-1)(1/2x^2+sqrt(1-x^2)1=(x^2)/4)=cos^(-1)x/2-cos^(-1)xdot`

Answer» `cos^(-1) ((1)/(2) x^(2) + sqrt(1 -x^(2)) sqrt(1 - (x^(2))/(4))) = cos^(-1) (x.(x)/(2) + sqrt(1 -x^(2)) sqrt(1 - ((x)/(2))^(2)))`
for `cos^(-1) ((1)/(2) x^(2) + sqrt(1 - x^(2)) sqrt(1 - (x^(2))/(4))) = cos^(-1) (x)/(2) - cos^(-1) x`
L.H.S. `gt 0`, hence R.H.S. `gt 0`
`rArr cos^(-1).(x)/(2) - cos^(-1) gt 0 " or " cos^(-1).(x)/(2) gt cos^(-1) x`
Since `cos^(-1) x` is a decreasing function, we get
`(x)/(2) le x rArr (x)/(2) ge 0 rArr x ge 0 rArr x in [0,1]`
Discussion
33.

Solve `sin^(-1) x + sin^(-1) (1 - x) = cos^(-1) x`

Answer» We have `sin^(-1) x + sin^(-1) (1 - x) = cos^(-1) x`
`rArr sin^(-1) [x sqrt(1 - (1-x)^(2)) + sqrt(1 - x^(2)) (1 -x)] = sin^(-1) sqrt(1 -x^(2))`
`rArr x sqrt(1-(1 - x)^(2)) + sqrt(1 - x^(2)) (1 - x) = sqrt(1 -x^(2))`
`rArr x sqrt(1-(1 - x)^(2)) = x sqrt(1 - x^(2))`
`rArr x = 0 " or " 2x - x^(2) = 1 - x^(2)`
`rArr x = 0 " or " x = (1)/(2)`
Discussion
34.

Solve `sin^(-1) x cos^(-1) x = sin^(-1) (3x -2)`

Answer» `sin^(-1) x - cos^(-1) x = sin^(-1) (3x -2)`
This equation is valid if `-1 le x le 1 and -1 le 3x - 2 le 1`
`rArr x in [(1)/(3), 1]`
given equation can be rewritten as:
`(pi)/(2) - cos^(-1) x = cos^(-1) x = (pi)/(2) - cos^(-1) (3x -2)`
`rArr 2 cos^(-1) x = cos^(-1) (3 x -2)`
`rArr cos^(-1) (2x^(2) - 1) = cos^(-1) (3x -2)`
`rArr 2x^(2) -1 = 3x - 2`
`rArr 2x^(2) - 3x + 1 = 0`
`rArr x = 1 " or " (1)/(2)`
Discussion
35.

If `f(x) = sin^(-1) x` then prove that `lim_(x rarr (1^(+))/(2)) f(3x -4x^(3)) = pi - 3 lim_(x rarr (1^(+))/(2)) sin^(-1) x`

Answer» `sin^(-1) (3x -4x^(3)) = pi - 3 sin^(-1) x " if " (1)/(2) lt x lt 1`
`:. underset(x rarr (1)/(2))("lim") f(3x - 4x^(3)) = underset(x rarr (1)/(2))("lim") (pi - 3 sin^(-1) x)`
`= pi - 3 underset(x rarr (1)/(2))("lim") sin^(-1) x`
Discussion
36.

solve the following equation `sec^(-1).(x)/(a) - sec^(-1).(x)/(b) = sec^(-1) b - sec^(-1) a, a ge 1, b ge 1, a!= b`

Answer» Correct Answer - `x = ab`
`sec^(-1).(x)/(a) - sec^(-1).(x)/(b) = sec^(-1) b - sec^(-1) a`
`rArr cos^(-1).(a)/(x) - cos^(-1).(b)/(x) = cos^(-1).(1)/(b) - cos^(-1).(1)/(a)`
`rArr cos^(-1).(a)/(x) + cos^(-1).(1)/(a) = cos^(-1).(b)/(x) - cos^(-1).(1)/(b)`
`rArr cos^(-1) [(1)/(x) - sqrt(1 - (a^(2))/(x^(2))) sqrt(1 - (1)/(a^(2)))] = cos^(-1) [(1)/(x) - sqrt(1 - (b^(2))/(x^(2))) sqrt(1 -(1)/(b^(2)))]`
`rArr (1)/(x) - sqrt(1 - (1)/(a^(2)) - (a^(2))/(x^(2)) + (1)/(x^(2))) = (1)/(x) - sqrt(1 - (b^(2))/(x^(2)) - (1)/(b^(2)) + (1)/(x^(2)))`
`rArr sqrt(1 -(1)/(a^(2)) - (a^(2))/(x^(2)) + (1)/(x^(2))) = sqrt(1 - (1)/(b^(2)) - (b^(2))/(x^(2)) + (1)/(x^(2)))`
`rArr -(1)/(a^(2)) - (a^(2))/(x^(2)) = -(1)/(b^(2)) - (b^(2))/(x^(2))`
`rArr (1)/(b^(2)) - (1)/(a^(2)) = (a^(2) - b^(2))/(x^(2))`
`rArr x^(2) = a^(2) b^(2)`
`:. x = +- ab`
But `x = 0 ab` does not satisfy the given equation
Hence, `x = ab` is the required solution
Discussion
37.

Let `a = cos^(-1) cos 20, b = cos^(-1) cos 30 and c = sin^(-1) sin (a + b)` then The largest integer x for which `sin^(-1) (sin x) ge |x -(a + b + c)|` isA. 1B. 2C. 3D. 4

Answer» Correct Answer - C
`a = 20 - 6pi, b = 10 pi - 30, c = sin^(-1) sin(4pi - 10) = 10 - 3pi`
So, `a + b + c = pi`
`sin^(-1) sin x ge |x -pi| rArr x in [(pi)/(2), pi]`
So, largest integer x = 3
Discussion
38.

Let `a = cos^(-1) cos 20, b = cos^(-1) cos 30 and c = sin^(-1) sin (a + b)` then If `5 sec^(-1) x + 10 sin^(-1) y = 10 (a + b + c)` then the value of `tan^(-1) x + cos^(-1) (y -1)` isA. `(pi)/(2)`B. `(pi)/(4)`C. `pi`D. 0

Answer» Correct Answer - B
`5 sec^(-1) x + 10 sin^(-1) y = 10 pi`
`rArr sec^(-1) x = pi and sin^(-1) y = (pi)/(2)`
`rArr x = - 1, y = 1`
So, `tan^(-1) (-1) + cos^(-1)(0) = -(pi)/(4) + (pi)/(2) = (pi)/(4)`
Discussion
39.

If `0lexle1` then `tan{1/2sin^(-1)((2x)/(1+x^(2)))+1/2cos^(-1)((1-x^(2))/(1+x^(2)))}` is equal toA. `(2x)/(1-x^(2))`B. 0C. `(2x)/(1+x^(2))`D. x

Answer» We know that
`sin^(-1)(2x)/(1+x^(2))=2tan^(-1)xif -1 lexle1`
and
`cos^(-1)(1+x^(2))/(1+x^(2))=2tan^(-1)x if 0lexleinfty`
`therefore tan{1/2sin^(-1)(2x)/(1+x^(2))+1/2cos^(-1)(1-x^(2))/(1+x^(2))}`
`=tan(tan^(-1)(2x)/(1-x^(2)))=(2x)/(1-x^(2))`
Discussion
40.

Consider the function `f(x) = sin^(-1)x`, having principal value branch `[(pi)/(2), (3pi)/(2)] and g(x) = cos^(-1)x`, having principal value brach `[0, pi]` The value of `f(sin 10)` isA. `10 - 3pi`B. `10 -2 pi`C. `10 -(5pi)/(2)`D. `(7pi)/(2) - 10`

Answer» Correct Answer - B
We have `f(x) = sin^(-1)x`, having range `[(pi)/(2), (3pi)/(2)]`
Consider the function `h(x) = (pi - sin^(-1) x)`, where
`sin^(-1) x in [-(pi)/(2), (pi)/(2)]`
Clearly, function `f(x) and h(x)` are identical.
Advantage with function `y = h(x)` is that we are well versed with `sin^(-1) x` if `(sin^(-1) x) in [-(pi)/(2), (pi)/(2)]`
`h(sin 10) = pi - sin^(-1) (sin 10)`
`= pi - sin^(-1) (sin (3pi - 10)))`
`= pi - (3pi - 10)`
`= 10 - 2pi`
Now, `f(x) lt (3pi)/(4)`
`:. h(x) lt (3pi)/(4)`
`rArr pi - sin^(-1) x lt (3pi)/(4)`
`rArr sin^(-1) x gt (pi)/(4)`
`rArr (pi)/(4) lt sin^(-1) x le (pi)/(2)`
`rArr (1)/(sqrt2) lt x le 1`
Discussion
41.

Find the domain of `sec^(-1)(2x+1)`.A. RB. [-1,1]C. `(-oo,-1] cup[0,oo)`D. `[-oo,-1]cup[1,oo)`

Answer» The domain of `sic^(-1)x is (-infty,-1]cup[1,infty)`
Therefore `sec^(-1)(2x+1)`is meaningful if
`2x+1ge1 or 2x+1le1`
`rarr 2xge0 or 2xle-2`
`rarr xge0 or xle-1`
`rarrx(-infty,-1]cup[0,infty)`
Hence the domain of `sec^(-1)(2x+1)is(-infty,-1]cup[0,infty)`
Discussion
42.

Find the principalvalue of `tan^(-1){sin(-pi/2)}`A. `(pi)/(4)`B. `-(pi)/(4)`C. `(3pi)/(4)`D. `(3pi)/(4)`

Answer» We know that sin`-(pi)/(2)=-1`
`therefore tan^(-1){sin(-pi)/(2)}=tan^(-1)(-1)=-(pi)/(2)`
Discussion
43.

The value of cot`[sin^(-1){cos(tan^(-1)1)}]` is

Answer» We know that `tan^(-1)1=(pi)/(4)`
`therefore cot[sin^(-1){cos(tan^(-1)1)}]`
`=cot{sin^(-1)(cos(pi)/(4))}=cot(sin^(-1))(1)/sqrt(2)=cot(pi)/(4)=1`
Discussion
44.

Evaluate `2tan^(-1)(1/2)+tan^(-1)(1/4)`

Answer» Correct Answer - `tan^(-1).(19/8)`
Discussion
45.

Prove that `tan^(- 1)(1/3)+tan^(- 1)(1/7)+tan^(- 1)(1/13)+..........+tan^-1 (1/(n^2+n+1))+......oo =pi/4`A. `(pi)/(2)`B. `(pi)/(4)`C. `(2pi)/(3)`D. 0

Answer» We have
`tan^(-1)1/3+tan^(-1)1/7+tan^(-1)1/13+..+tan^(-1)(1)/(n^(2)+n_+1)+…to infty`
`=underset(nrarrinfty)lim underset(r=1)overset(n)Sigma tan^(-1){(1)/(1+r(r+1))}`
`=underset(nrarrinfty)lim underset(r=1)oveset(n)Sigma tan^(-1){(r+1)-r)/(1+r(r+1))`
`=underset(nrarrinfty)lim (tan^(-1)(n+1)-tan^(-1)1)`
`=tan^(-1)infty-tan^(-1)=(pi)/(12)-(pi)/(4)-(pi)/(4)`
Discussion
46.

Find the value of: `tan^(-1)[2cos(2sin^(-1)(1/2)]`A. `(pi)/(4)`B. `-(pi)/(4)`C. `(3pi)/(4)`D. `(3pi)/(4)`

Answer» We know that `sin^(-1)1/2=(pi)/(6)`
`therefore tan^(-1){2cos(2sin^(-1)1/2)}`
`=tan^(-1)(2xx1/2)=tan^(-1)1=(pi)/(4)`
Discussion
47.

Simplity `tan{1/2sin^(-1)((2x)/(1+x^(2)))+1/2cos^(-1)(1-y^(2))/(1+y^(2))}`, if `xgtygt1`.

Answer» Correct Answer - `(x+y)/(1-xy)`
Discussion
48.

If `tan^(-1)((x+1)/(x-1))+tan^(-1)((x-1)/(x))=tan^(-1)(-7)+pi` then x =A. 2B. 3C. 4D. none of these

Answer» We know that
`tan^(-1)x+tan^(-1)y=pi+tn^(-1)(x+y)/(1-xy)`
If `xgt0ygt0and xy gt1`
`therefore tan^(-1)(x+1)/(x-1)+tan^(-1)(x-1)/(x-1)xx(x-1)/(x)gt0`
`and ((x+1)/(x-1)+(x-1)/(x))/(1-(x+1)/(x-1)xx(x-1)/(x)=-7`
`rarr x in (-infty,-1)cup(1,infty) and (2x^(2)-x+1)/(1-x)=-7`
`rarr x in(-infty,-1)cup(1,infty) and (x-2)^(2)=0`
`rarr` x=2
Discussion
49.

If `0lexle1` then `sin{tan^(-1)((1-x^(2))/(2x))+cos^(-1)((1-x^(2))/(1+x^(2)))}` is equal toA. 1B. `-1`C. 0D. none of these

Answer» We know that
`tan^(-1)((2x)/(1-x^(2))=2 tan^(-1)x if -1 ltx le1`
and
`therefore sin{ tan^(-1)(1-x^(2))/(2x)+cos^(-1)(1-x^(2))/(1+x^(2))}`
`=sin{cot^(-1)(2x)/(1-x^(2))+cos^(-1)(1-x^(2))/(1+x^(2))`
`=sin{(pi)/(2)-2tan^(-1) x+2 tan^(-1)x} if 0 le x lt1`
`=sin(pi)/(2)=1`
Discussion
50.

If `-1lexle0` then `tan{1/2sin^(-1)((2x)/(1+x^(2)))+1/2cos^(-1)((1-x^(2))/(1+x^(2)))}` is equal toA. `(2x)/(1-x^(2))`B. 0C. `(2x)/(1+x^(2))`D. x

Answer» We know that
`sin^(-1)(2x)/(1+x^(2))=-2 tan^(-1) x if -1 le x le 1`
and
` cos^(-1)((1-x^(2))/(1+x^(2)))=-2 tan^(-1)x if -oo lex le 0`
`therefore tan{1/2sin^(-1)(2x)/(1+x^(2))+1/2 cos^(-1)(1-x^(2))/(1+x^(2)`
`=tan(tan^(-1)x-tan^(-1)x) if -1 le x le 0 =-tan 0=0`
Discussion